An object initially at rest suddenly explodes in two fragments of masses $2.6 \mathrm {~kg}$ and $3.3 \mathrm {~kg}$ that move in opposite directions. If the speed of the first fragment is $3.6 \mathrm {~m} / \mathrm { s } \text {, }$ find the internal energy of the explosion.

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Quizplus · Accepted Answer

The internal energy of the explosion is equal to the kinetic energy of the fragments. The kinetic energy of the first fragment is:$$K_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (2.6 \mathrm{~kg}) (3.6 \mathrm{~m/s})^2 = 18.48 \mathrm{~J}$$The kinetic energy of the second fragment is:$$K_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (3.3 \mathrm{~kg}) v_2^2$$Since the fragments move in opposite directions, the total kinetic energy is:$$K = K_1 + K_2 = 18.48 \mathrm{~J} + \frac{1}{2} (3.3 \mathrm{~kg}) v_2^2$$The internal energy of the explosion is:$$U = K = 18.48 \mathrm{~J} + \frac{1}{2} (3.3 \mathrm{~kg}) v_2^2$$We don't know the value of $v_2$, but we can use the conservation of momentum to find it:$$m_1 v_1 = m_2 v_2$$$$v_2 = \frac{m_1 v_1}{m_2} = \frac{(2.6 \mathrm{~kg})(3.6 \mathrm{~m/s})}{3.3 \mathrm{~kg}} = 2.7 \mathrm{~m/s}$$Substituting this into the equation for the internal energy, we get:$$U = 18.48 \mathrm{~J} + \frac{1}{2} (3.3 \mathrm{~kg}) (2.7 \mathrm{~m/s})^2 = 30 \mathrm{~J}$$Therefore, the internal energy of the explosion is 30 J.

An Object Initially at Rest Suddenly Explodes in Two Fragments 2.6 kg2.6 \mathrm {~kg}2.6 kg

An Object Initially at Rest Suddenly Explodes in Two Fragments $2.6 \mathrm {~kg}$